When electronic transition occurs from higher energy state to lower energy state with energy difference equal to `Delta E` electron volts , the wavelength of the line emitted is approximately equal to

Options

A. `(12395)/(Delta E ) xx 10^(-10)m`
B. `(12395)/(Delta E ) xx 10^(10)m`
C. `(12395)/(Delta E ) xx 10^(-10)m`
D. `(12395)/(Delta E ) xx 10^(10)m`

Explanation

To solve the problem of finding the wavelength of light emitted when an electron transitions from a higher energy state to a lower energy state, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Transition**: When an electron transitions from a higher energy state (n2) to a lower energy state (n1), it emits a photon. The energy of this photon corresponds to the difference in energy levels, denoted as ΔE (in electron volts).

2. **Convert ΔE to Joules**: The energy difference ΔE is given in electron volts (eV). To convert this to joules (J), we use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Therefore,
\[ \Delta E \text{ (in J)} = \Delta E \text{ (in eV)} \times 1.6 \times 10^{-19} \]

3. **Use Planck's Equation**: According to Planck's quantum theory, the energy of a photon is related to its wavelength (λ) by the equation:
\[ \Delta E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \text{ J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \text{ m/s} \)),
- \( \lambda \) is the wavelength in meters.

4. **Rearranging for Wavelength**: We can rearrange the equation to solve for the wavelength:
\[ \lambda = \frac{hc}{\Delta E} \]

5. **Substituting Values**: Substitute the values of \( h \), \( c \), and \( \Delta E \) (converted to joules) into the equation:
\[ \lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^{8} \text{ m/s})}{\Delta E \times (1.6 \times 10^{-19})} \]

6. **Simplifying the Expression**: This can be simplified further:
\[ \lambda = \frac{1.989 \times 10^{-25}}{\Delta E \times (1.6 \times 10^{-19})} \]
\[ \lambda \approx \frac{12.43 \times 10^{-7}}{\Delta E} \text{ meters} \]
This gives us the approximate wavelength in meters.

7. **Final Result**: The wavelength of the light emitted is approximately:
\[ \lambda \approx \frac{12.43 \times 10^{-7}}{\Delta E} \text{ m} \]